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Let $$Lu:=-\sum_{i,j=1}^n(a^{i,j}(t,x)u_{x_i}(t,x))_{x_j}+\sum_{i=1}^nb^i(t,x)u_{x_i}(t,x)+c(t,x)u(t,x)$$ and the associated bilinear form $$B[u,v;t]:=\int_U\sum_{i,j=1}^na^{i,j}(t,x)u_{x_i}(t,x)v_{x_j}(t,x)+\sum_{i=1}^nb^i(t,x)u_{x_i}(t,x)v(t,x)+c(t,x)u(t,x)v(t,x)dx$$. The notation is from Evans "Partial Differential Equation". Now suppose that I know that $u_m$ converges weakly to $u$ in the space $L^2(0,T;H^1_0(U))$, where $U$ is nice (open, bounded, etc.). Weak convergence in this space means

$$\lim_{m\to \infty}\int_0^T\langle \phi,u_m\rangle dt= \int_0^T\langle \phi,u\rangle dt$$ for all $\phi\in L^2(0,T;H^{-1}(U))$. I know that for every $\phi\in H^{-1}(U)$ there exists $\phi^0,\dots,\phi^n\in L^2(U)$ such that $\langle \phi,u\rangle = \int_U \phi^0u+\sum_{i=1}^n \phi^iu_{x_i}dx$ for all $u\in H^1_0(U)$.

Now my question is, why does $\int_0^TB[u_m,v;t]dt$ converges to $\int_0^TB[u,v;t]$? Of course, we have to use the weak convergence, but I do not see how exactly.

Just for completeness: for a Banach space $X$, we define $L^2(0,T;X)$ as the space of all measurable function $f:[0,T]\to X$ such that $\sqrt{\int_0^T\|f\|^2_Xdt}<\infty$.

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Hi! I don't really understand how $B[u_m,v;t]$ can converge "weakly" in $L^2(0,T;H_0^1(U))$. However, I think one can show that it converges in $L^2(0,T)$ because $B$ is continuous for all $t$ (the duality product should converge pointwise a.e. because of the weak convergence) – vanguard2k Aug 23 '12 at 14:43
@vanguard2k Your right! I edited my question. Sorry for the mistake! Can you turn your comment into an answer? Because I got stuck at writing it down correctly. – user20869 Aug 23 '12 at 14:51
up vote 5 down vote accepted

Here's a sketch:

  1. Show that the bilinear forms $B[\cdot,\cdot\;; t]$ are continuous on $H^1_0(U)$, i.e. there is a constant $C$ such that $|B[f,g;t]| \le C \|f\|_{H^1_0(U)} \|g\|_{H^1_0(U)}$. Moreover, the constant can be taken independent of $t$. (Use Cauchy-Schwarz repeatedly. Here I'm assuming that the coefficient functions $a^{i,j}, b^i, c$ are bounded.)

  2. Conclude that the bilinear form $(u,v) \mapsto \int_0^T B[u,v;t]\,dt$ is continuous on $L^2(0,T; H^1_0(U))$. (Cauchy-Schwarz will be used again.)

  3. In particular, $u \mapsto \int_0^T B[u,v; t]\,dt$ is a continuous linear functional on $L^2(0,T; H^1_0(U))$.

  4. Recall the definition of weak convergence in a Banach space: $x_n \to x$ weakly iff $\ell(x_n) \to \ell(x)$ for every continuous linear functional $f$. (The "definition" you give is based on the fact that the dual space of $L^2(0,T;H^1_0(U))$ can be naturally identified with $L^2(0,T; H^{-1}(U))$.)

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I've really got nothing to add. So there is no need to turn my comment into an answer. +1 – vanguard2k Aug 23 '12 at 16:31
Thank you so much for this satisfying answer! – user20869 Aug 24 '12 at 6:04

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